Fe2o3 + 6hcl -> 2fecl3 + 3h2o (1)
nFe2o3 = m/M=8/160=0.05mol
theo phương trình 1 : nHcl=6nFe2o3=0.05.6=0.3 mol
nFecl3=2nFe2o3=0.05.2=0.1mol
nH2o=3nFe2o3=0.05.3=0.15mol
=> vHcl=n/Cm=0.3/1=0.3 lít
mFecl3=n.M=0.1.162.5=16.25g
mH2o=n.M=0.15.18=2.7g
=>mdd=mFecl3+mH2o=16.25+2.7=18.95g
=>c%=mFecl3/mdd.100%=16.25/18.95.100%=85.75%